By Ellina Grigorieva

This e-book goals to dispel the secret and worry skilled via scholars surrounding sequences, sequence, convergence, and their functions. the writer, an comprehensive girl mathematician, achieves this by way of taking an issue fixing technique, beginning with attention-grabbing difficulties and fixing them step-by-step with transparent causes and illuminating diagrams.

The reader will locate the issues attention-grabbing, strange, and enjoyable, but solved with the rigor anticipated in a contest. a few difficulties are taken without delay from arithmetic competitions, with the identify and 12 months of the examination supplied for reference. facts thoughts are emphasised, with quite a few equipment provided. The textual content goals to extend the brain of the reader by means of usually proposing a number of how you can assault an identical challenge, in addition to drawing connections with diversified fields of arithmetic. Intuitive and visible arguments are provided along technical proofs to supply a well-rounded methodology.

With approximately three hundred difficulties together with tricks, solutions, and options, *Methods of fixing Sequences and sequence Problems* is a perfect source for these studying calculus, getting ready for arithmetic competitions, or simply trying to find a beneficial problem. it may even be utilized by school who're trying to find fascinating and insightful difficulties that aren't normally present in different textbooks.

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**Extra resources for Methods of Solving Sequence and Series Problems**

**Example text**

B1 Á r nÀ1 þ . . ð1:16Þ If the number of terms is infinitely great (n approaches infinity), then the expression on the right of Eq. 16 is called an infinite geometric series. Consider an infinite 1Àrn Þ decreasing geometric series. The partial sum is Sn ¼ b1 ð1Àr . If jrj < 1andn ! 1, t hen r n ! 1 b1 : 1Àr ð1:17Þ We notice that the sequence of partial sums has a limit. However, if jrj > 1 by Eq. 15 the sequence of the partial sums will dramatically increase without bound. In order to illustrate this phenomenon, let us solve Problem 17.

2 À Á 1 À 21n 1 1 n Finally, we get Sn ¼ 1 Á À n Á n ¼ 2 À nÀ1 À n . 4 Finding the nth Term of a Sequence or Series 37 1 Answer. un ¼ 2nn , Sn ¼ 2 À 2nÀ1 À 2nn , S1 ¼ 2: What if each term of a sequence is a partial sum of an arithmetic progression? Then a1 ¼ S 1 , S1 ¼ b1 a2 ¼ S 2 , S2 ¼ b1 þ b2 , b2 ¼ b1 þ 1 Á d a3 ¼ S 3 , S3 ¼ b1 þ b2 þ b3 , b3 ¼ b1 þ 2 Á d ... an ¼ Sn , Sn ¼ b1 þ b2 þ b3 þ . . þ bn , bn ¼ b1 þ ðn À 1Þ Á d: Let us examine this by solving Problem 32. Problem 32 Find the nth term of a sequence 3, 13, 30, 54, 85, 123, .

An ¼ n2 : On the other hand, a sequence of the squares is a sequence of partial sums for the arithmetic sequence: 1, 3, 5, 7, 9, . . and the following is true: a1 ¼ 12 ¼ 1 a2 ¼ 22 ¼ 1 þ 3 a3 ¼ 32 ¼ 1 þ 3 þ 5 a4 ¼ 42 ¼ 1 þ 3 þ 5 þ 7 a5 ¼ 52 ¼ 1 þ 3 þ 5 þ 7 þ 9 ... an ¼ n2 ¼ 1 þ 3 þ 5 þ . . 4. 4 Each square can be represented as a sum of n consecutive odd numbers, starting from 1. Ancient Greeks called this a sequence of square numbers and tried to prove everything geometrically. For example, they knew that if one takes one stone and three stones, then one can make a square with side two.